javax.servlet.UnavailableException：缺少路径/WEB-INF/struts-config.xml的配置资源

时间：2013-11-09 11:06:31

标签： java jsp web-applications struts

我是Struts的新手，我正在尝试使用Struts运行示例登录Web应用程序，但由于(javax.servlet.UnavailableException: Missing configuration resource for path /WEB-INF/struts-config.xml)错误，我无法获得结果。我已经完成了在eclipse中配置struts的所有工作。我甚至不知道什么例外？请任何人帮助我

我得到的错误是： 

javax.servlet.UnavailableException: Missing configuration resource for path /WEB-INF/struts-config.xml
at org.apache.struts.action.ActionServlet.splitAndResolvePaths(ActionServlet.java:1872)
at org.apache.struts.action.ActionServlet.initModuleConfig(ActionServlet.java:683)
at org.apache.struts.action.ActionServlet.init(ActionServlet.java:356)
at javax.servlet.GenericServlet.init(GenericServlet.java:212)
at org.apache.catalina.core.StandardWrapper.loadServlet(StandardWrapper.java:1213)
at org.apache.catalina.core.StandardWrapper.load(StandardWrapper.java:1026)
at org.apache.catalina.core.StandardContext.loadOnStartup(StandardContext.java:4421)
at org.apache.catalina.core.StandardContext.start(StandardContext.java:4734)
at org.apache.catalina.core.ContainerBase.start(ContainerBase.java:1057)
at org.apache.catalina.core.StandardHost.start(StandardHost.java:840)
at org.apache.catalina.core.ContainerBase.start(ContainerBase.java:1057)
at org.apache.catalina.core.StandardEngine.start(StandardEngine.java:463)
at org.apache.catalina.core.StandardService.start(StandardService.java:525)
at org.apache.catalina.core.StandardServer.start(StandardServer.java:754)
at org.apache.catalina.startup.Catalina.start(Catalina.java:595)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
at java.lang.reflect.Method.invoke(Unknown Source)
at org.apache.catalina.startup.Bootstrap.start(Bootstrap.java:289)
at org.apache.catalina.startup.Bootstrap.main(Bootstrap.java:414)

index.jsp 

<html> 
    <body>


        <form action="login.do">
       <tr>
       <td>Username</td> 
       <td><input type = "text" name = "username" maxlength = "25"></td>
       </tr>
       <tr>
       <td>Password</td>
       <td><input type = "password" name = "password" minlength = "8" maxlength = "16"></td>
       </tr>

       <tr>
       <td  colspan = "2" align = "center"><button type = "submit" name = "submit">Submit</button>
       <button type = "reset"   name = "reset">Reset</button></td>

       </tr>


        </form>
    </body>
</html>

LoginForm.java 

package com.sample.com;

import org.apache.struts.action.ActionForm;

import org.apache.struts.action.Action;

import org.apache.struts.action.ActionMapping;

import org.apache.struts.action.ActionForward;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;

public class LoginAction extends Action {

    public ActionForward execute(ActionMapping mapping, ActionForm form, HttpServletRequest req, HttpServletResponse res)
    {
        System.out.println("test");

    LoginForm login = (LoginForm)form;

    String uname=login.getUname();

    String pass=login.getPass();

    if(uname=="admin" && pass =="test")
    {
    return mapping.findForward("success");

    }
    else
    {
    return mapping.findForward("failure");

    }
  }
}

LoginAction.java 

package com.sample.com;

import org.apache.struts.action.ActionForm;

import org.apache.struts.action.Action;

import org.apache.struts.action.ActionMapping;

import org.apache.struts.action.ActionForward;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;



public class LoginAction extends Action {

    public ActionForward execute(ActionMapping mapping, ActionForm form, HttpServletRequest req, HttpServletResponse res)
    {

        System.out.println("test");

    LoginForm login = (LoginForm)form;

    String uname=login.getUname();

    String pass=login.getPass();

    if(uname=="admin" && pass =="test")
    {
    return mapping.findForward("success");
    }
    else
    {
    return mapping.findForward("failure");

    }

  }


}

config-struts.xml 

<?xml version="1.0" encoding="UTF-8" ?>



<struts-config>

    <form-beans>
        <form-bean name="loginRequest" type="com.sample.com.LoginForm"/>
    </form-beans> 

    <global-forwards>
        <forward name="success"  path="/success.jsp"/>
    </global-forwards>

    <action-mappings>
        <action path="/login" type="com.sample.com.LoginAction" name="loginRequest">
            <forward name="success" path="/success.jsp"/>
            <forward name="failure" path="/failure.jsp"/>        
        </action>
    </action-mappings>

</struts-config>

web.xml 

<?xml version="1.0" encoding="UTF-8"?>

<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">

    <servlet>
        <servlet-name>action</servlet-name>
        <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
        <init-param>
            <param-name>config</param-name>
            <param-value>/WEB-INF/struts-config.xml</param-value>
        </init-param>
        <init-param>
            <param-name>debug</param-name>
            <param-value>2</param-value>
        </init-param>
        <init-param>
            <param-name>detail</param-name>
            <param-value>2</param-value>
        </init-param>
        <load-on-startup>2</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>action</servlet-name>
        <url-pattern>*.do</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>
</web-app>

4 个答案:

答案 0 :(得分：2)

确保struts-config.xml位于WEB-INF文件夹中。同样，hibernate.cfg.xml也应该在适当的位置（如果您正在使用它）。

答案 1 :(得分：1)

您已将Web应用程序配置为使用struts-config.xml 

<init-param>
    <param-name>config</param-name>
    <param-value>/WEB-INF/struts-config.xml</param-value>
</init-param>

但是在指定的文件夹中没有具有此类名称的文件。确保该文件存在，或者如果您对struts配置文件使用其他名称，则可以更改此init参数值以反映文件可用性。如果模块中有多个配置文件，则可以使用逗号分隔名称。

答案 2 :(得分：1)

您需要从apache lib 文件夹中添加 servlet-api.jar 。您需要在java资源下的 src 文件夹中包含 struts.xml 文件夹。

答案 3 :(得分：0)

1）您已将struts.xml文件名称为 config-struts.xml ，但在您编写为 struts-config.xml 的init-param中。请正确检查名称。

2）或者也许（常见问题），因为你缺少struts xml文件中的DTD标记。尝试将此代码添加到 struts-config.xml 中的<struts-config>标记上方。

<!DOCTYPE struts-config PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 1.3//EN" "http://struts.apache.org/dtds/struts-config_1_3.dtd">

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